Question

5 cos square 60 degree + 4 second square 30 degree minus 10 square 45 degree / sin square 30 degree + cos 30 degree​

Answer 1

Answer:

(5 COS² 60 + 4 SEC²30 - TAN²45)/ (SIN²30+COS²30)

===> 5(1/2)² + 4(2/√3)² - (1)² / (1)

===> 5/4 + 4(4/3) - 1

===> 5/4 + 16/3 - 1

==> (15+64-12)/12

==> 67/12

Answer 2

$\dfrac{5\cos^260^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\cos^2 30^\circ}=\dfrac{67}{12}$

Step-by-step explanation:

Given : Expression $\dfrac{5\cos^260^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\cos^2 30^\circ}$

To find : Evaluate the expression ?

Solution :

Using trigonometric values,

$\cos 60^\circ=\frac{1}{2}\\\sec 30^\circ=\frac{2}{\sqrt3}\\\tan 45^\circ =1\\\sin 30^\circ=\frac{1}{2}\\\cos 30^\circ=\frac{\sqrt3}{2}$

Substitute the values,

$=\dfrac{5(\frac{1}{2})^2+4(\frac{2}{\sqrt3})^2-(1)^2}{(\frac{1}{2})^2+(\frac{\sqrt3}{2})^2}$

$=\dfrac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

$=\dfrac{\frac{15+64-12}{12}}{\frac{4}{4}}$

$=\dfrac{\frac{67}{12}}{1}$

$=\dfrac{67}{12}$

Therefore, $\dfrac{5\cos^260^\circ+4\sec^2 30^\circ-\tan^2 45^\circ}{\sin^2 30^\circ+\cos^2 30^\circ}=\dfrac{67}{12}$

#Learn more

EVALUATE : SIN THETA + COS THETA + SIN THETA COS (90°- THETA) COS THETA / SEC(90°-theta) + cos theta sin (90-theta)sin theta /cosec(90°-theta) - 2 sin(90°-theta) cos (90°-theta)

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